Chapter 21, Problem 62A

(a)

Interpretation Introduction

Interpretation: The half-reactions for the given process need to be determined.

Concept Introduction: In an electrochemical process, a redox reaction takes place. Here, two half-reactions occur where in one reaction oxidation takes place and in the other reaction, reduction takes place. The sum of these two-half reactions gives the overall reaction for the process.

In oxidation, the loss of electrons takes place by an element and its oxidation state increases Similarly in reduction, the gain of electrons takes place by an element. Here, the oxidation state of the element decreases.

Explanation of Solution

According to the process given in the question, aluminum metal is produced when ore containing aluminum oxide is converted into aluminum chloride. Here, aluminum metal and chlorine gas are produced by the electrolysis of molten aluminum chloride.

At the anode, oxidation takes place and at the cathode, reduction takes place thus, the possible two half-reactions will be:

Anode-Oxidation reaction:

  $2{\text{Cl}}^{-}\to {\text{Cl}}_2+2e$

Cathode-Reduction reaction:

  ${\text{Al}}^{3+}+3e\to \text{Al}$

(b)

Interpretation Introduction

Interpretation: The overall cell reaction needs to be determined.

Concept Introduction: In an electrochemical process, a redox reaction takes place. Here, two half-reactions occur where in one reaction oxidation takes place and in the other reaction, reduction takes place. The sum of these two-half reactions gives the overall reaction for the process.

In oxidation, the loss of electrons takes place by an element and its oxidation state increases Similarly in reduction, the gain of electrons takes place by an element. Here, the oxidation state of the element decreases.

Explanation of Solution

The overall cell reaction is obtained by adding two half-reactions. This is represented as follows:

  $\begin{array}{l}3\left(2{\text{Cl}}^{-}\to {\text{Cl}}_2+2e\right)\\ 2\left({\text{Al}}^{3+}+3e\to \text{Al}\right)\\ \overline{\underset{¯}{2{\text{Al}}^{3+}+6{\text{Cl}}^{-}+2\text{Al}+3{\text{Cl}}_2}}\end{array}$

Therefore, the overall cell reaction is as follows:

  $2{\text{Al}}^{3+}+6{\text{Cl}}^{-}+2\text{Al}+3{\text{Cl}}_2$

(c)

Interpretation Introduction

Interpretation: The products obtained at the cathode and the anode need to be identified.

Concept Introduction: In an electrochemical process, a redox reaction takes place. Here, two half-reactions occur where in one reaction oxidation takes place and in the other reaction, reduction takes place. The sum of these two-half reactions gives the overall reaction for the process.

In oxidation, the loss of electrons takes place by an element and its oxidation state increases Similarly in reduction, the gain of electrons takes place by an element. Here, the oxidation state of the element decreases.

Explanation of Solution

The two half-reactions are as follows:

Anode-Oxidation reaction:

  $2{\text{Cl}}^{-}\to {\text{Cl}}_2+2e$

Cathode-Reduction reaction:

  ${\text{Al}}^{3+}+3e\to \text{Al}$

Thus, the products obtained at the cathode and anode are Al and ${\text{Cl}}_{\text{2}}$ respectively.